Question

我目前有这些PHP页面，它允许我向数据库添加记录。（在这种情况下是它的成员）它在我可以添加，删除和查看的意义上完美地工作。但我不知道如何获得编辑（或UPDATE功能。

这是我的数据库连接代码：

<?php

// Server Info
$server = 'localhost';
$username = 'root';
$password = '';
$database = 'gamgam';

// Connect to database
$connection = new mysqli($server, $username, $password, $database);

?>

这是我的添加代码：

<!DOCTYPE html>
<html>
<head><title>Insert Users</title></head>
<body>
<h2>Insert User Confirmation</h2>

<form action="<?php $_SERVER['PHP_SELF']?>" method="post"/> <br> 
<?php

require_once('connection.php');

echo    "<label for='memberID' >Member ID:</label>";
echo    "<input type='text' name='memberID' id='memberID' />";
echo    "<br /><br />"; 

echo    "<label for='username' >Username:</label>";
echo    "<input type='text' name='username' id='username' />";
echo    "<br /><br />"; 

echo    "<label for='password' >Password:</label>";
echo    "<input type='password' name='password' id='password' />";
echo    "<br /><br />"; 

echo    "<label for='fName' >Firstname:</label>";
echo    "<input type='text' name='fName' id='fName' />";
echo    "<br /><br />"; 

echo    "<label for='lName' >Lastname:</label>";
echo    "<input type='text' name='lName' id='lName'  />";
echo    "<br /><br />"; 

echo    "<label for='address' >Address:</label>";
echo    "<input type='text' name='address' id='address'  />";
echo    "<br /><br />"; 

echo    "<label for='email' >Email:</label>";
echo    "<input type='text' name='email' id='email'  />";
echo    "<br /><br />";

echo    "<input type='submit' name='submit' value='Submit' />";
echo    "<input type='reset' value='Clear' />";
echo    "<br /><br />"; 
?>
</form>
</section>

<p><a href='login.php'>Login</a></p>    

<?php
if(!isset($_POST['submit'])) {  
    echo 'Please Register';
} else {

    $memberID = $_POST['memberID'];
    $username = $_POST['username'];
    $password = $_POST['password'];
    $fName = $_POST['fName'];
    $lName = $_POST['lName'];
    $address = $_POST['address'];
    $email = $_POST['email'];


    $query = "INSERT INTO `members` 
                  (MemberID, Username, Password, FirstName, LastName, 
                   StreetAddress, Email) 
                VALUES ('$memberID', '$username', '$password', '$fName', 
                        '$lName', '$address', '$email')";
        mysqli_query($connection, $query)
        or  die(mysqli_error($connection));

        $rc = mysqli_affected_rows($connection);
        if ($rc==1)   
        {
            echo '<h4>The database has been updated with the following details: </h4> ';
            echo 'MemberID: '.$memberID.'<br />';
            echo 'Username: '.$username.'<br />';
            echo 'Password: '.$password.'<br />';
            echo 'First Name: '.$fName.'<br />';
            echo 'Last Name: '.$lName.'<br />';
            echo 'Address: '.$address.'<br />';
            echo 'Email: '.$email.'<br />';
        } else {    
            echo '<p>The data was not entered into the database this time.</p>';   
        }
    }
?>
</body>
</html>

这是我的观看代码：

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html>
<head>
    <title>View Records</title>
</head>
<body>

<table border="1" style="width:100%" >

<?php
/* 
    VIEW.PHP
    Displays all data from 'players' table
*/

    // connect to the database
    include('connection.php');

    // get results from database
    $result = mysqli_query($connection, "SELECT * FROM members") 
        or die(mysqli_error());  

    // loop through results of database query, displaying them in the table
    while($row = mysqli_fetch_array( $result )) {

        // echo out the contents of each row into a table
        echo "<tr>";
        echo '<td>' . $row['MemberID'] . '</td>';
        echo '<td>' . $row['Username'] . '</td>';
        echo '<td>' . $row['Password'] . '</td>';
        echo '<td>' . $row['FirstName'] . '</td>';
        echo '<td>' . $row['StreetAddress'] . '</td>';
        echo '<td>' . $row['Email'] . '</td>';

        echo '<td><a href="edit.php?MemberID=' . $row['MemberID'] . '">Edit</a></td>';
        echo '<td><a href="delete.php?MemberID=' . $row['MemberID'] . '">Delete</a></td>';
        echo "</tr>"; 
    } 

    // close table>
    echo "</table>";
?>
<p><a href="insert_user.php">Add a new record</a></p>

</body>
</html>

这是删除代码：

<?php

    // Connect to the database
    include('connection.php');

    // Confirm that the 'code' variable has been set
    if (isset($_GET['MemberID']))
    {
        // Get the 'MemberID' variable from the URL
        $MemberID = $_GET['MemberID'];

        // Delete record from database
        if ($stmt = $connection->prepare("DELETE FROM members WHERE MemberID = ? LIMIT 1")) {
            $stmt->bind_param("i",$MemberID);   
            $stmt->execute();
            $stmt->close();

        } else {
            echo "ERROR: could not prepare SQL statement.";
        }
        $connection->close();

        // Redirect user after delete is successful
        header("Location: view.php");
    } else {
    // If the 'code' variable isn't set, redirect the user
        header("Location: view.php");
    }

?>

我在网上经历了许多基本的php表单模板，试图将他们所做的工作融入到实现结果中但没有取得任何成功。需要为我的网站编写哪些代码才能具有编辑已在数据库中创建的记录的功能，而无需通过phpmyadmin。任何帮助都是相关的。

Answer 1

Edit就像Add一样，但您需要先读取记录并填充字段值。

从add的代码开始，执行以下操作：

<?php $MemberID = (int) $_GET['MemberID']; ?>
<form action="<?php $_SERVER['PHP_SELF']?>" method="post"/> <br> 
<input type="hidden" name="MemberID" value="<?php echo $MemberID; ?>"
<?php
require_once('connection.php');
$result = mysqli_query($connection, "SELECT * FROM members where MemberID = $MemberID") or die(mysqli_error());  

// loop through results of database query, displaying them in the table
$row = mysqli_fetch_assoc($result);
extract($row);

echo    "<label for='memberID' >Member ID:</label>";
echo    "$memberID"; // member ID should not be editable
echo    "<br /><br />"; 

echo    "<label for='username' >Username:</label>";
echo    "<input type='text' name='username' id='username' value="$username" />";
echo    "<br /><br />";

PHP代码将有一个像

这样的查询

`UPDATE `members` SET `username` = '$username' ...  WHERE `MemberID` = '$MemberID'"

编辑记录PHP表单

1 个答案: